If $m(d)=Fd$ and $l(d)=\mathbf{1}^T d$ then what is $\nabla_m (l)$?

Question

Given a vector $d\in\mathbb{R}^p$, matrix $F\in\mathbb{R}^{q\times p}$, and unit vector $\mathbf{1}\in\mathbb{R}^{p}$, suppose the vector- and scalar-valued functions $m$ and $l$ are defined as $m(d)=Fd$ and $l(d)=\mathbf{1}^Td$ respectively. Is there a way to calculate the closed form of $\nabla_m (l)$? Also, is $\nabla_m (l)$ constant on account of $l$ and $m$ being linear functions of $d$?

Answer 1

I’ll assume that the columns of $F$ are independent. Then there exists a $p \times q$ matrix $G$ (e.g. the pseudo inverse) such that $GF = 1$. Now

$$l(d) = {\bf1}^{\mathrm t} d = {\bf1}^{\mathrm t} Gm = (G^{\mathrm t} {\bf1})^{\mathrm t} m.$$ This shows that $\nabla_m l = G^{\mathrm t} {\bf 1}$. Note that even though $G$ might not be unique, this gradient is well-defined on the column space $F\mathbb R^p$ (the domain of $m$).

Answer 2

As an addendum to WimC's answer, I'll add that under certain conditions, the gradient is constant even if $F$ is a 'fat' matrix.

Assumption: Assume that $p>q$ and $\mathbf{1}_p \in \text{Range}(F^\top)$.

Claim: $\nabla_m(l)$ is constant.

Proof: There exists $r\in\mathbb{R}^q$ such that $\mathbf{1}_p = F^\top r$. Thus $l = \mathbf{1}^\top d = r^\top Fd = r^\top m$, and it follows that $\nabla_m(l) = r = \text{constant}$.